Is ${730397}$ divisible by $3$ ?
A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {730397}= &&{7}\cdot100000+ \\&&{3}\cdot10000+ \\&&{0}\cdot1000+ \\&&{3}\cdot100+ \\&&{9}\cdot10+ \\&&{7}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {730397}= &&{7}(99999+1)+ \\&&{3}(9999+1)+ \\&&{0}(999+1)+ \\&&{3}(99+1)+ \\&&{9}(9+1)+ \\&&{7} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {730397}= &&\gray{7\cdot99999}+ \\&&\gray{3\cdot9999}+ \\&&\gray{0\cdot999}+ \\&&\gray{3\cdot99}+ \\&&\gray{9\cdot9}+ \\&& {7}+{3}+{0}+{3}+{9}+{7} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first five terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${730397}$ is divisible by $3$ if ${ 7}+{3}+{0}+{3}+{9}+{7}$ is divisible by $3$ Add the digits of ${730397}$ $ {7}+{3}+{0}+{3}+{9}+{7} = {29} $ If ${29}$ is divisible by $3$ , then ${730397}$ must also be divisible by $3$ ${29}$ is not divisible by $3$, therefore ${730397}$ must not be divisible by $3$.